Question: You have found the following ages (in years) of all 6 gorillas at your local zoo: $ 2,\enspace 1,\enspace 13,\enspace 6,\enspace 4,\enspace 15$ What is the average age of the gorillas at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{2 + 1 + 13 + 6 + 4 + 15}{{6}} = {6.8\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $-4.8$ years $23.04$ years $^2$ $1$ year $-5.8$ years $33.64$ years $^2$ $13$ years $6.2$ years $38.44$ years $^2$ $6$ years $-0.8$ years $0.64$ years $^2$ $4$ years $-2.8$ years $7.84$ years $^2$ $15$ years $8.2$ years $67.24$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{23.04} + {33.64} + {38.44} + {0.64} + {7.84} + {67.24}} {{6}} $ $ {\sigma^2} = \dfrac{{170.84}}{{6}} = {28.47\text{ years}^2} $ The average gorilla at the zoo is 6.8 years old. The population variance is 28.47 years $^2$.